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Is there a clever way to determine a primitive element of the finite extension $$F=\mathbb{Q}(\sqrt{2}+i,\sqrt{3}-i)/\mathbb{Q} \text{ ?}$$ On simpler examples, I've been able to find one by determining all field morphisms $\sigma: F\to\mathbb{C}$ such that $\sigma|_\mathbb{Q}=\text{id}$ and finding $x\in F$ with a different image under each such $\sigma$. But here, it seems quite painful since the degree is between 8 and 16...

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The degree is at most $8$, since the field is contained in $\mathbb{Q}(\sqrt{2},\sqrt{3},i)$, which has degree $8$ over $\mathbb{Q}$. – Arturo Magidin Jun 20 '12 at 18:21
up vote 5 down vote accepted

It is relatively straightforward to show that $F=\mathbb{Q}(i,\sqrt2,\sqrt3)$, because it is easy to see that $\mathbb{Q}(\sqrt2+i)=\mathbb{Q}(\sqrt2,i)$, and similarly $\mathbb{Q}(\sqrt3-i)=\mathbb{Q}(\sqrt3,i)$. Thus $[F:\mathbb{Q}]=8$, this extension is Galois, and the Galois group is elementary 2-abelian. An automorphism $\sigma$ is uniquely determined once we specify $\sigma(i)=\pm i$ (two choices), $\sigma(\sqrt2)=\pm\sqrt2$ and $\sigma(\sqrt3)=\pm\sqrt3$ (again two choices). If an element $z\in F$ is not fixed by any of these automorphisms, it is not in any of the subfields, so for example $$ z=2i+3\sqrt2+5\sqrt3 $$ will be a primitive element.

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Did you choose these coefficients on $i, \sqrt2, \sqrt3$ for any particular reason? – Dylan Moreland Jun 20 '12 at 18:45
No :-) ${}{}{}{}$ – Jyrki Lahtonen Jun 20 '12 at 18:46
Oh, I didn't remark that $F=\mathbb{Q}(i,\sqrt{2},\sqrt{3})$. Then it's much easier, thanks! – Klaus Jun 20 '12 at 18:51
@Klaus: It is easier than checking that $z$ is not in any of the 14 intermediate fields, and much easier than showing directly that we can write the generators of $F$ as rational expressions of $z$. – Jyrki Lahtonen Jun 20 '12 at 18:55

The Galois group of $\mathbb{Q}(\sqrt{2},\sqrt{3},i)$ over $\mathbb{Q}$ is of order $8$; it should not be hard to verify that it is in fact isomorphic to $C_2\times C_2\times C_2$; the generators are complex conjugation, the map that sends $\sqrt{2}$ to $-\sqrt{2}$ and fixes $\sqrt{3}$ and $i$; and the map that sends $\sqrt{3}$ to $-\sqrt{3}$ and fixes $\sqrt{2}$ and $i$.

Now, your field contains $\sqrt{2}+\sqrt{3}+i$. It has eight different images under the automorphisms of $\mathbb{Q}(\sqrt{2},\sqrt{3},i)$: $\pm\sqrt{2}\pm\sqrt{3}\pm i$. So $\mathbb{Q}(\sqrt{2}+\sqrt{3}+i) = \mathbb{Q}(\sqrt{2},\sqrt{3},i)$. Also, your field is contained in $\mathbb{Q}(\sqrt{2},\sqrt{3},i)$, hence in $\mathbb{Q}(\sqrt{2}+\sqrt{3}+i)$.

Your field contains $\sqrt{2}+\sqrt{3}$, and so contains $\sqrt{2}$ and $\sqrt{3}$ (since $\mathbb{Q}(\sqrt{2}+\sqrt{3}) = \mathbb{Q}(\sqrt{2},\sqrt{3})$). So it is equal to $\mathbb{Q}(\sqrt{2},\sqrt{3},i)$, and you are done.

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